Locally Euclidean topological space

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• I am currently doing the proofs for equivalent definitions Alec (talk) 16:55, 19 February 2017 (UTC)

Caveat:Need to do locally euclidean of dimension $n$!

Definition

Let $(X,\mathcal{ J })$ be a topological space, we say it is locally Euclidean if:

• $\forall p\in X\exists n\in\mathbb{N}_0\exists U\in\mathcal{O}(p;X)\underbrace{\exists V\in\mathcal{O}(\mathbb{R}^n)\exists\varphi\in\mathcal{F}(U,V)[U\cong_\varphi V]}_{\exists\text{homeomorphism }\varphi:U\rightarrow\text{some open subset of }\mathbb{R}^n }$
  • In words: for all points, $p\in X$, there is an $n$ such that there is an open neighbourhood of $p$ which is homeomorphic to some open set in $\mathbb{R}^n$ for the given {{M|n}].

Equivalent definitions

Some open ball at the origin

Claim:

• $\Big(\forall p\in X\exists n\in\mathbb{N}_0\exists U\in\mathcal{O}(p;X)\exists V\in\mathcal{O}(\mathbb{R}^n)\exists\varphi\in\mathcal{F}(U,V)[U\cong_\varphi V]\Big)$ $\iff$ $\Big(\forall p\in X\exists n\in\mathbb{N}_0\exists U\in\mathcal{O}(p;X)\exists\epsilon\in\mathbb{R}_{>0}\exists\varphi\in\mathcal{F}(U,B_\epsilon(0;\mathbb{R}^n)[U\cong_\varphi B_\epsilon(0;\mathbb{R}^n)]\Big)$

Proof:

$\implies$

• Let $p\in X$ be given
  • Choose $n:\eq n'$ where $n'$ is the $n\in\mathbb{N}_0$ posited to exist by the LHS of the $\iff$
    • We now obtain:
      • $U'\in\mathcal{O}(p;X)$ posited to exist by the LHS,
        • $V'\in\mathcal{O}(\mathbb{R}^n)$ posited to exist by the LHS, such that:
          • $\varphi':U'\rightarrow V'$ posited to exist by the LHS is a homeomorphism.
    • Recall that "an open set in a metric space contains an open ball about all of its points", this means:
      • $\exists\delta\in\mathbb{R}_{>0}[B_\delta(\varphi'(p);\mathbb{R}^n)\subseteq V']$
    • As open balls are open sets and "the pre-image of an open set under a homeomorphism is open" we see:
      • $\varphi'^{-1}\big(B_\delta(\varphi'(p);\mathbb{R}^n)\big)$ is open in $X$
    • We must now show that $p\in\varphi'^{-1}\big(B_\delta(\varphi'(p);\mathbb{R}^n)\big)$ (so we can say $\varphi'^{-1}\big(B_\delta(\varphi'(p);\mathbb{R}^n)\big)\in\mathcal{O}(p,X)$ shortly)
      • Note that $f(p)\in B_\delta(\varphi'(p);\mathbb{R}^n)$ as $d(\varphi'(p),\varphi'(p)):\eq 0$ regardless of the metric used
        • As $0<\delta$ we see $\varphi'(p)\in B_\delta(\varphi'(p);\mathbb{R}^n)$
          • Thus $p\in\varphi'^{-1}\big(B_\delta(\varphi'(p);\mathbb{R}^n)\big)$ - by the definition of pre-image
    • Choose: $U:\eq \varphi'^{-1}\big(B_\delta(\varphi'(p);\mathbb{R}^n)\big)$, by the discussion above we see $U\in\mathcal{O}(p,X)$
      • Choose: $\epsilon\eq\delta$, we found $\delta$ above, recall $\delta\in\mathbb{R}_{>0}$, so obviously $\epsilon\in\mathbb{R}_{>0}$

Grade: A

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The bulk of the proof is done here, the "not trivial part" is that $\varphi'$ restricts to a homeomorphism onto $B_\delta(\varphi'(p);\mathbb{R}^n)$, then compose that with a translation, we use translations are homeomorphisms and we're basically done. Alec (talk) 16:55, 19 February 2017 (UTC)

• "Given a homeomorphism all subspaces of the domain are homeomorphic to their image under the homeomorphism itself" will be needed and is ready